A long straight wire of radius 10mm carries a current uniformly distributed over its cross section. The ration of magnetic fields due to the wire at distance of 5mm and 15 mm, respectively from the axis of the wire is

Options

A. `3/4`
B. `2/3`
C. `1/2`
D. `2`

Explanation

To find the ratio of the magnetic fields due to a long straight wire at distances of 5 mm and 15 mm from the axis of the wire, we can follow these steps:

### Step 1: Understand the Magnetic Field due to a Long Straight Wire
The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying a current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 \) is the permeability of free space.

### Step 2: Calculate the Magnetic Field at 15 mm (B2)
For the distance of 15 mm (which is 0.015 m):
\[ B_2 = \frac{\mu_0 I}{2 \pi (0.015)} \]

### Step 3: Calculate the Magnetic Field at 5 mm (B1)
For the distance of 5 mm (which is 0.005 m):
\[ B_1 = \frac{\mu_0 I}{2 \pi (0.005)} \]

### Step 4: Find the Ratio of B1 to B2
Now, we need to find the ratio \( \frac{B_1}{B_2} \):
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2 \pi (0.005)}}{\frac{\mu_0 I}{2 \pi (0.015)}} \]

### Step 5: Simplify the Ratio
The \( \mu_0 \) and \( I \) terms cancel out:
\[ \frac{B_1}{B_2} = \frac{0.015}{0.005} = 3 \]

### Final Answer
Thus, the ratio of the magnetic fields at distances of 5 mm and 15 mm from the wire is:
\[ \frac{B_1}{B_2} = 3 \]